Question

The radii of the ends of a frustum of a cone 45 cm high are 28 cm and 7 cm. Find its volume, the curved surface area and the total suface area (Take $\pi$= $\frac{22}{7}$))

Answer 1

The frustum can be viewed as a difference of two right circular cones OAB and OCD (see Fig.).

Let the height (in cm) of the cone OAB be $h_1$and its slant height $l_1$, i.e., $OP=h_1$ and $OA=OB=l_1$.

Let $h_2$ be the height of cone OCD and l2 its slant height. We have: $r_1=28cm,r_2=7cmandtheheightoffrustum\left(h\right)=45cm.$

Also, $h_1=45+h_2$ (1)

We first need to determine the respective heights $h_{1}and{h}_2$, of the cone OAB and OCD. Since the triangles OPB and OQD are similar (Why?), we have $\frac{h_1}{h_2}=\frac{28}{7}=\frac{4}{1}$ (2)

From (1) and (2), we get $h_2=15and{h}_1=60$.

Now, the volume of the frustum = volume of the cone OAB - volume of the cone OCD = $[\frac{1}{3}.\frac{22}{7}.(28{)}^2.\left(60\right)-\frac{1}{3}.\frac{22}{7}.(7{)}^2.(15\left)\right]c{m}^3=48510c{m}^3$

The respective slant height $l_2$and $l_1$of the cones OCD and OAB are given by $l_2=\sqrt{(7{)}^2+(15{)}^2}=16.55cm\left(approx\right)$ $l_1=\sqrt{(28{)}^2+(60{)}^2}=4\sqrt{(7{)}^2+(15{)}^2}=4\times 16.55=66.20cm$

Thus , the curved surface area of the frustum = $\pi r_1{l}_1-\pi r_2{l}_2$ = $\frac{22}{7}\left(28\right)\left(66.20\right)-\frac{22}{7}\left(7\right)\left(16.55\right)=5461.5c{m}^2$

Now, the total surface area of the frustum $=thecurvedsurfacearea+\pi {r_1}^2+\pi {r_2}^2$ $=5461.5c{m}^2+\frac{22}{7}(28{)}^{2}c{m}^2+\frac{22}{7}(7{)}^{2}c{m}^2$ $=5461.5c{m}^2+2464c{m}^2+154c{m}^2=8079.5c{m}^2$.

Let h be the height, l the slant height and $r_{1}and{r}_2$ the radii of the ends $(r_1>r_2)$ of the frustum of a cone.

Then we can directly find the volume, the curved surface area and the total surface area of frustum by using the formulae given below

(i) Volume of the frustum of the cone = $\frac{1}{3}\pi h({r_1}^2+{r_2}^2+r_1{r}_2)$ .

(ii) The curved surface area of the frustum of the cone = $\pi (r_1+r_2)l$ where l = $\sqrt{h^2(r_1{r}_2{)}^2}$ .

(iii) Total surface area of the frustum of the cone = $\pi l(r_1+r)2)+\pi {r_1}^2+\pi {r_2}^2.$ where l = $\sqrt{h^2(r_1{r}_2{)}^2}$ . These formulae can be derived using the idea of similarity of triangles but we shall not be doing derivations here,

Let us solve Example 12, using these formulae: (i) volume of the frustum = $\frac{1}{3}\pi h({r_1}^2+{r_2}^2+r_1{r}_2)$ = $\frac{1}{3}.\frac{22}{7}.45.\left[\right(28{)}^2+(7{)}^2+(28\left)\right(7\left)\right]c{m}^3$= 48510 $c{m}^3$

(ii) We have l = $\sqrt{h^2+(r_1+r_2{)}^2}=\sqrt{(45{)}^2+(28-7{)}^2}$ cm = 3 $\sqrt{(15{)}^2+(7{)}^2}$ = 49.65 cm So, the curved surface area of the frustum = $\pi (r_1+r_2)l=\frac{22}{7}(28+7)\left(4.65\right)=5461.5c{m}^2$

(iii) Total curved surface area of the frustum = $\pi (r_1+r_2)l+\pi {r_1}^2+\pi {r_2}^2$ = $[5461.5+\frac{22}{7}(28{)}^2+\frac{22}{7}\left(7{)}^2\right]c{m}^2=8079.5c{m}^2$